The thing is that when you are talking about the Lagrangian $\mathcal{L}$ it is to be considered as a function of the fields. So it is $\mathcal{L}[A_\mu,\partial_\nu A_\mu]$. One considers all components of $A$, that is $A_\mu$ for $\mu=0,1,2,3$ and all derivatives of all components of $A$, that is $\partial_\nu A_\mu$ for $\mu,\nu=0,1,2,3$ to be independent variables.
Imagine that you are dealing with a function of several variables where the variables are the components of $A$ and the derivatives.
So the derivative
$$\dfrac{\partial}{\partial(\partial_\nu A_\mu)}$$
Is a derivative with respect to the particular coordinate $\partial_\nu A_\mu$ for this $\mu,\nu$.
Remember that for independent coordinates $x^\mu$ the equation
$$\dfrac{\partial x^\mu}{\partial x^\nu}=\delta^\mu_\nu$$
holds. The same happens here. Since $A_\mu$ and $\partial_\alpha A_\beta$ are independent coordinates we have
$$\dfrac{\partial A_\mu}{\partial(\partial_\alpha A_\beta)}=\dfrac{\partial (\partial_\alpha A_\beta)}{\partial A_\mu}=0,\quad \dfrac{\partial A_\mu}{\partial A_\nu}=\delta_{\mu\nu},\quad \dfrac{\partial (\partial_\alpha A_\beta)}{\partial (\partial _\mu A_\nu)}=\delta_{\alpha\mu}\delta_{\beta\nu}.$$
Convince yourself of that, considering the equation above well known in $\mathbb{R}^n$. It is the same equation with just different names for the coordinates!
To deal with the ones with raised index just explicitly write the metric. So we know that $\partial^\mu A^\nu = g^{\mu\lambda}g^{\nu\sigma}\partial_\lambda A_\sigma$.
In particular we have
$$\dfrac{\partial( \partial^\mu A^\nu)}{\partial(\partial_\alpha A_\beta)}=\dfrac{\partial( g^{\mu\sigma}g^{\nu\rho} \partial_\sigma A_\rho)}{\partial(\partial_\alpha A_\beta)}=g^{\mu\sigma}g^{\nu\rho}\dfrac{\partial( \partial_\sigma A_\rho)}{\partial(\partial_\alpha A_\beta)}=g^{\mu\sigma}g^{\nu\rho} \delta_{\sigma\alpha}\delta_{\rho\beta}=g^{\mu\alpha}g^{\nu\beta}.$$
Then just differentiate $\mathcal{L}$ normaly as function of these coordinates. We have
$$\mathcal{L}=-\dfrac{1}{2}(\partial^\mu A^\nu\partial_\mu A_\nu-\partial^\mu A^\nu \partial_\nu A_\mu)$$
To not mix the indices from the contraction with the free index from the derivative you want to perform, use a different one. We will differentiate with respect to $\partial_\alpha A_\beta$
$$\dfrac{\partial \mathcal{L}}{\partial(\partial_\alpha A_\beta)}=-\dfrac{1}{2}\left(\dfrac{\partial (\partial^\mu A^\nu)}{\partial(\partial_\alpha A_\beta)}\partial_\mu A_\nu+\partial^\mu A^\nu \dfrac{\partial(\partial_\mu A_\nu)}{\partial(\partial_\alpha A_\beta)}-\dfrac{\partial(\partial^\mu A^\nu)}{\partial(\partial_\alpha A_\beta)}\partial_\nu A_\mu - \partial^\mu A^\nu \dfrac{\partial(\partial_\nu A_\mu)}{\partial(\partial_\alpha A_\beta)}\right)$$
Now using what is above the result follows.
